Question

Regular gasoline averaged ​$2.83 per gallon in December 2018. Assume the standard deviation for gasoline prices is ​$0.14 per gallon. A random sample of 40 service stations was selected. Complete parts a through d.

a. What is the probability that the sample mean will be less than ​$2.84?

The probability that the sample mean will be less than $2.84 is ______

​(Type an integer or decimal rounded to four decimal places as​ needed.)

b. What is the probability that the sample mean will be more than ​$2.87?

The probability that the sample mean will be more than ​$2.87 is ________.

​(Type an integer or decimal rounded to four decimal places as​ needed.)

c. What is the probability that the sample mean will be between $2.81 and $2.91​?

The probability that the sample mean will be between $2.81 and ​$2.91 is _________.

​(Type an integer or decimal rounded to four decimal places as​ needed.)

d. Suppose the sample mean is $2.89. Does this result support the average gasoline price​ findings? Explain your answer.

The probability that the average price per gallon is more than ​$2.89 is ________.

​(Type an integer or decimal rounded to four decimal places as​ needed.)

Does this result support the average gasoline price findings? Explain your answer. Consider a probability of less than 0.05 to be small.

A) The probability supports the finding that the average price per gallon for gas in the population is ​$2.83​, because this probability is small.

B) The probability does not support the finding that the average price per gallon for gas in the population is ​$2.83​, because this probability is large. "

C)The probability supports the finding that the average price per gallon for gas in the population is ​$2.83, because this probability is large.

D)The probability does not support the finding that the average price per gallon for gas in the population is ​$2.83​, because this probability is small.

Answer 1

Part a)
X ~ N ( µ = 2.83 , σ = 0.14 )
P ( X < 2.84 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 2.84 - 2.83 ) / ( 0.14 / √40 )
Z = 0.4518
P ( ( X - µ ) / ( σ/√(n)) < ( 2.84 - 2.83 ) / ( 0.14 / √(40) )
P ( X < 2.84 ) = P ( Z < 0.45 )
P ( X̅ < 2.84 ) = 0.6743

Part b)
X ~ N ( µ = 2.83 , σ = 0.14 )
P ( X > 2.87 ) = 1 - P ( X < 2.87 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 2.87 - 2.83 ) / ( 0.14 / √ ( 40 ) )
Z = 1.807
P ( ( X - µ ) / ( σ / √ (n)) > ( 2.87 - 2.83 ) / ( 0.14 / √(40) )
P ( Z > 1.81 )
P ( X̅ > 2.87 ) = 1 - P ( Z < 1.81 )
P ( X̅ > 2.87 ) = 1 - 0.9646
P ( X̅ > 2.87 ) = 0.0354

Part c)
X ~ N ( µ = 2.83 , σ = 0.14 )
P ( 2.81 < X < 2.91 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 2.81 - 2.83 ) / ( 0.14 / √(40))
Z = -0.9035
Z = ( 2.91 - 2.83 ) / ( 0.14 / √(40))
Z = 3.614
P ( -0.9 < Z < 3.61 )
P ( 2.81 < X̅ < 2.91 ) = P ( Z < 3.61 ) - P ( Z < -0.9 )
P ( 2.81 < X̅ < 2.91 ) = 0.9998 - 0.1831
P ( 2.81 < X̅ < 2.91 ) = 0.8167

Part d)
X ~ N ( µ = 2.83 , σ = 0.14 )
P ( X > 2.89 ) = 1 - P ( X < 2.89 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 2.89 - 2.83 ) / ( 0.14 / √ ( 40 ) )
Z = 2.7105
P ( ( X - µ ) / ( σ / √ (n)) > ( 2.89 - 2.83 ) / ( 0.14 / √(40) )
P ( Z > 2.71 )
P ( X̅ > 2.89 ) = 1 - P ( Z < 2.71 )
P ( X̅ > 2.89 ) = 1 - 0.9966
P ( X̅ > 2.89 ) = 0.0034

D)The probability does not support the finding that the average price per gallon for gas in the population is ​$2.83​, because this probability is small.